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  • Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released?

Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury T = 435.5 \times 10^{-3} \frac{N}{m}.

Answers (1)

Surface tension energy, E=\sigma \Delta A

As we can infer from the law of conservation of mass, the volume of a drop V = v1 + v2

r_1 = 0.1 cm = 0.001 m

r_2 = 0.2 cm = 0.002 m

\Delta A = 4 \pi r_1^{2} + 4 \pi r_2^{2} - 4\pi R^{2}

\Delta A = 4 \pi \left [r_1^{2} + r_2^{2}- R^{2} \right ]

The new bigger drop has a radius of R,

\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi \left [ r_1^{3}+r_2^{3} \right ]

R^{3}= r_1^{3}+r_2^{3}

R^{3}= 9 \times 10^{-9}

R^{3}= 2.1 \times 10^{-3}m

Now, E=\sigma \Delta A

E = 4 \times 3.14 [(10^{-3})^{2} + (2 \times 10^{-3})^{2} - (2.1 \times 10^{-3})^{2}] \times 435.5 \times 10^{-3}

E = 1742 \times 3.14 \times 10^{-9} [0.59] = 5469.88 \times 0.59 \times 10^{-9}

E = 32.27 \times 10^{-7} J

Finally, the area will be smaller, when energy is released as the bigger drop is formed from smaller drops.

Posted by

infoexpert24

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